Arithmetic derivative: Difference between revisions
Added C
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;*[[wp:Arithmetic_derivative|Wikipedia: Arithmetic Derivative]]
=={{header|C}}==
{{trans|Go}}
<syntaxhighlight lang="c">#include <stdio.h>
#include <stdint.h>
typedef uint64_t u64;
void primeFactors(u64 n, u64 *factors, int *length) {
if (n < 2) return;
int count = 0;
int inc[8] = {4, 2, 4, 2, 4, 6, 2, 6};
while (!(n%2)) {
factors[count++] = 2;
n /= 2;
}
while (!(n%3)) {
factors[count++] = 3;
n /= 3;
}
while (!(n%5)) {
factors[count++] = 5;
n /= 5;
}
for (u64 k = 7, i = 0; k*k <= n; ) {
if (!(n%k)) {
factors[count++] = k;
n /= k;
} else {
k += inc[i];
i = (i + 1) % 8;
}
}
if (n > 1) {
factors[count++] = n;
}
*length = count;
}
double D(double n) {
if (n < 0) return -D(-n);
if (n < 2) return 0;
int i, length;
double d;
u64 f[80], g;
if (n < 1e19) {
primeFactors((u64)n, f, &length);
} else {
g = (u64)(n / 100);
primeFactors(g, f, &length);
f[length+1] = f[length] = 2;
f[length+3] = f[length+2] = 5;
length += 4;
}
if (length == 1) return 1;
if (length == 2) return (double)(f[0] + f[1]);
d = n / (double)f[0];
return D(d) * (double)f[0] + d;
}
int main() {
u64 ad[200];
int n, m;
double pow;
for (n = -99; n < 101; ++n) {
ad[n+99] = (int)D((double)n);
}
for (n = 0; n < 200; ++n) {
printf("%4ld ", ad[n]);
if (!((n+1)%10)) printf("\n");
}
printf("\n");
pow = 1;
for (m = 1; m < 21; ++m) {
pow *= 10;
printf("D(10^%-2d) / 7 = %.0f\n", m, D(pow)/7);
}
return 0;
}</syntaxhighlight>
{{out}}
<pre>
As Go example
</pre>
=={{header|C++}}==
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